Question 553689
<pre>
Ellipses do not have asymptotes.  Hyperbolas do.  Did you mean 
"hyperbola" and typed "ellipse" instead?  I will assume that's
what you did.

Since one asymptote is y = {{{1/4}}}x, the othe one is y = {{{-1/4}}}x.

We draw those, plot the vertices (9,0) and (-9,0), and and sketch the hyperbola to approach them.

{{{drawing(800,800/3,-15, 15,-5,5, graph(800,800/3,-15,15,-5,5,sqrt(x^2-81)/4),

graph(800,800/3,-15,15,-5,5,-sqrt(x^2-81)/4),green(line(16,4,-16,-4), line(16,-4,-16,4)) )}}}

Then we draw in the defining rectangle:

  {{{drawing(800,800/3,-15, 15,-5,5, graph(800,800/3,-15,15,-5,5,sqrt(x^2-81)/4),green(rectangle(-9,-9/4,9,9/4),line(-9,9/4,9,9/4),line(-9,-9/4,9,-9/4)),

graph(800,800/3,-15,15,-5,5,-sqrt(x^2-81)/4),green(line(16,4,-16,-4), line(16,-4,-16,4)) )}}}

By letting x=4 and -4 in each of those asymptote equations we can 
see that the corners of the defining rectangle are:

(9,{{{9/4}}}), (9,{{{-9/4}}}), (-9,{{{-9/4}}}), and (-9,{{{9/4}}})

so a = 9 and b = {{{9/4}}}, so the equation is

{{{x^2/9^2}}} - {{{y^2/(9/2)^2}}} = 1

Edwin</pre>